February 17, Simplex Method Continued
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1 February 17, 2005 Simplex Method Continued 1
2 Today s Lecture Review of the simplex algorithm. Formalizing the approach Alternative Optimal Solutions Obtaining an initial bfs Is the simplex algorithm finite? (Answer, yes, but only if we are careful) 2
3 The simplex algorithm (for max problems) Start with a feasible corner point solution find an improved corner point solution No Is it optimal? No Yes quit with optimal solution Is the optimum unbounded from above? Yes quit with proof of unboundedness 3
4 LP Canonical Form iff there is a basis that is feasible. z x 1 x 2 x 3 x = = 6 0` = 2 The basic variables are z, x 3 and x 4. The non-basic variables are x 1 and x 2. The basic feasible solution (bfs) is z = 0, x 1 = x 2 = 0, x 3 = 6, x 4 = 2 4
5 Optimality Conditions: maximization form z 1 0 x 1 x 2 x 3 x 4 -c c = = 6 The bar indicates that it is possibly the coefficient after some pivots = 2 The bfs is optimal if all coefficients z-row are 0. Use c j to denote the cost coefficients. Opt. conditions: The bfs is optimal if 0 for all j. -c j 5
6 The Simplex Pivot Rule z x 1 x 42 x 3 x 4 x = = = = 4 Pivot in a variable whose cost-row coefficient is negative. If there are multiple choices, then use a refined rule. If 0 for all j, then the basis is optimal. -c j 6
7 Review from last lecture: To do with your partner z x 1 x 42 x 3 x 4 x = = = = 4 1. Determine the basic variables and the basic solution 2. What will be the entering variable? 3. What will be the variable that leaves the basis? Reminder: set the value of the entering variable to. 7
8 The min ratio rule z x 1 x 42 x 3 x 4s x 45 If a is 0 for all i, then the objective value is unbounded from above c -2 s 0 = 3z a2 3s 1 = b a 1s -1 a 2s 0 0 = = b b 2 Min Ratio = min{ b i / a is : a js > 0 and i = 1 to m }. Usually, the argmin value is denoted as r. In this case, r = 3. Pivot on a rs 8
9 Minimum Ratio Rule Pivot out the basic variable in row r, where r = argmin i { b i / a is : a is > 0}, and thus b r / a rs = min { b i / a is : a is > 0}. If a is 0 for all i, then the solution is unbounded. 9
10 The pivot z x 1 x 42 x 3 x 4 x = /2 0 = /2 0 = /2 1 = The objective value strictly improves so long as b r > The new (basic feasible solution) bfs is feasible. 10
11 Alternative Optima (maximization) z x 1 x 2 x 3 x 4 Recall: z + 4 x 2 = 8 Note that z does not depend on x = = = 2 This basic feasible solution is optimal! There is an alternative optimum because the non-basic variable x 1 has a 0 cost-coefficient. If x 1 enters the basis, what variable leaves? 11
12 Alternative Optima (maximization) z x 1 x 2 x 3 x c = c = 6 = 2 There may be alternative optima if -c j 0 for all j and c j = 0 for some j where x j is non-basic Use the min ratio rule to determine which variable leaves the basis. 12
13 Alternative Optima (maximization) z x 1 x 2 x 3 x = = = 28 Perform the pivot. Note: the solution is different, but the objective value is the same. 13
14 Review of notation A basic feasible solution is optimal if - c j 0 for all j. Assumption: the entering variable is x s (and so - c s < 0) Pivot out the basic variable in row r, where r = argmin i { b i / a is : a is > 0}, and thus b r / a rs = min { b i / a is : a is > 0}. If a is 0 for all i, then the solution is unbounded. 14
15 Simplex Method (Max Form) Step 0. The problem is in canonical form and b 0. Step 1. If - c 0 then stop. The solution is optimal. If we continue, then there exists some - c j < 0. Step 2. Choose any non-basic variable to pivot in with - c s < 0, e.g., - c s = min j { - c j - c j < 0 }. If a is 0 for all i, then stop; the LP is unbounded. If we continue, then there exists some a is > 0. Step 3. Pivot out the basic variable in row r, where r is chosen by the min ratio rule, that is r = argmin i ( b i / a is : a is > 0 ). Step 4. Replace the basic variable in row r with variable x s and reestablish canonical form (i.e., pivot on the coefficient a rs. ) Step 5. Go to Step 1. 15
16 Preview of what is to come Obtaining an initial canonical form Degeneracy and improving solutions Proving finiteness and optimality under no degeneracy Handling degeneracy 16
17 The Phase 1 method to obtain an initial bfs We need an initial bfs. Approach: create a new LP that is related to the original LP and has the following features: 1. It is easy to find a bfs for the new LP 2. An optimal bfs for the new LP is a bfs for the original problem. 17
18 Step 0. Start with a tableau for the original LP z x 1 x 2 x 3 x = = = 2 Note: this is different than our running example from the last lecture. 18
19 Step 1. Write the constraints of the original LP x 1 x 2 x 3 x = = 2 FACT: Once we find a basic feasible solution, we can reconsider the original cost coefficients. 19
20 Step 2. Create new variables called artificial variables which will be the basic variables for the new LP. x 1 x 2 x 3 x 4 x 5 x = = 2 We will choose an objective function soon. 20
21 Step 3. Choose an objective so that the artificial variables will be 0 in an optimal solution, assuming that there is a bfs for the original problem. w x 1 x 2 x 3 x 4 x 5 x = = = 2 The objective function is w = -x 5 x 6 max w = 0 if and only if there is a feasible solution for the original problem We use w to not confuse it z. 21
22 Step 4. Subtract constraints 1, 2 (etc) to the objective constraint so that the tableau becomes canonical w x 1 x 2 x 3 x 4 x 5 x = = = 2 22
23 Step 5. Solve the Phase 1 Problem using the simplex algorithm. BV w x 1 x 2 x 3 x 4 x 5 x 6 w = 80 x /6 12 1/6 5 1/3 1-1/2 0 = 61 x /6 1 11/6 3 2/3 0-1/2 1 = 23 x 2 23
24 Step 6. Eliminate the artificial variables, and put back in the original objective function. BV wz x 1 x 2 x 3 x 4 x 5 x 6 wz = 0 x /6 1/6 1/3-1/2 = 1 x /6 11/6 2/3-1/2 = 3 24
25 Step 7. Add multiples of constraints (1) (2), etc to get the tableau in canonical form BV -w z x 1 x 2 x 3 x 4 x 5 x 6 -w z / / = 03 x /6 1/6 1/3-1/2 = 1 x 1 x /6 11/6 2/3-1/2 = 3 Subtract 3 times constraint 1 from the objective. Add 2 times constraint 2. 25
26 Step 8. Solve the original LP starting with the previous bfs BV -w z x 1 x 2 x 3 x 4 x 5 x 6 z = x 1 0 x /3-1/2 = /3-1/2 = This tableau was reached after one more pivot 26
27 Phase 1 method: a summary 1. Write the constraints of the original LP 2. Create new variables called artificial variables 3. New objective: Minimize the sum of the artificials 4. Put the tableau in canonical form 5. Solve the Phase 1 Problem 6. Write the old objective function in the optimal tableau for the phase 1 problem. (delete artificials too) 7. Put the tableau in canonical form 8. Solve the original LP starting with the feasible bfs that you have obtained. 27
28 More on the phase 1 method What happens if w < 0 at the end of phase 1? It means the original problem is infeasible Is it possible that w = 0 at the end of phase 1 and there are still artificial variables in the basis? Yes, for technical reasons; but it is possible to find a new bfs with w = 0 and without the artificial variables. So, there is no real problem. The phase 1 method is widely used. There are bells and whistles to speed it up. An alternative approach: the big M method 28
29 The Big M-Method Starts off the same as the Phase 1 approach. It creates a new problem with artificial variables. x 1 x 2 x 3 x 4 x 5 x = = 2 29
30 The Big M-Method, continued Choose original objective function but penalize artificial variables maximize z = -3x 1 + 2x 2 + 4x 3 + x 4-99x 5 99x 6 z x 1 x 2 x 3 x 4 x 5 x = = = 2 If the profit is sufficiently negative, then an optimal solution will set x 5 = x 6 = 0. 30
31 Big M Method continued. Then one gets the basis into canonical form, and pivots to obtain an optimal solution to the new problem, which will also be optimal for the original problem. If some artificial variable is in the basis at the end, then make its profit even more negative and continue. If an artificial variable is positive no matter what we do, then there was no feasible solution to the original problem. 31
32 Is the Simplex Algorithm Finite? The number of basic feasible solutions is at most n! / (n-m)! m!, which is the number of ways of selecting m basic variables out of n. It is finite if we can guarantee that a sequence of pivots cannot lead back to the same bfs It is finite if we can guarantee that each bfs has a strictly better objective than the last bfs. The only thing preventing objective values from improving is degeneracy. 32
33 An improving pivot BV z x 1 x 2 x 3 x 4 z = 2 z = x 1 = 0 x = 6 x 2 = x 3 = 6-3 x = 2 x 4 = 2-2 Suppose that variable x 2 enters the basis. = 1. The objective value will increase by 2. If the RHS is strictly positive, the objective will improve. 33
34 A non-improving (degenerate) pivot BV z x 1 x 2 x 3 x 4 z = 2 z = x 1 = 0 x = 6 x 2 = x 3 = 6-3 x = 20 x 4 = 0-2 Suppose that variable x 2 enters the basis. = 0. The solution stays the same. But the basis will change. 34
35 Degeneracy BV z x 1 x 2 x 3 x 4 z = 2 x = 6 x = 20 A bfs is degenerate if b j = 0 for some j. Otherwise, it is non-degenerate. This bfs is degenerate. In a degenerate pivot, the solution may stay the same. 35
36 A degenerate pivot BV z x 1 x 2 x 3 x 4 z = 2 x /2 0 = 6 x /2 1 = 20 0 The entering variable is x 2. The exiting variable is the one in constraint 2. In this case the solution did not change. (But the tableau did.) In a non-degenerate pivot, z 0 increases. 36
37 Theorem. If every basic feasible solution is non-degenerate, the simplex method is finite. Fact. If bases can be degenerate, it is possible that the simplex method will cycle, that is repeat bases in a periodic manner. 37
38 Overview on how to make the simplex method finite. Reduce it to a previously solved problem Replace the LP by a similar LP that has the following features: The RHS of the new problem is nearly the same as the original problem. It differs by an almost infinitesimal amount every feasible basis for the perturbed problem is also a feasible basis for the original problem an optimal basis for the perturbed problem is also an optimal basis the original problem 38
39 Creating the new LP BV z x 1 x 2 x 3 x 4 z = 0 x = 6 +ε x = 2 +ε 2 In theory ε must be absurdly small (less than ) for this technique to work. In practice, the simplex algorithm doesn t repeat bases, and so they don t worry about implementing this approach. Perturbations will not be on the quiz, exam, or on homeworks. 39
40 z x 1 x 2 x 3 x 4 1 a = b = c 0 0 = = 5 To do with your partner (4 minutes) Give conditions on a, b and c so that the following are true. 1. The current bfs is the unique optimal solution. 2. The current bfs is optimal, but there are alternative optima. 3. The current bfs is degenerate. 4. The current bfs is non-optimal, and x 3 is pivoted out at the next iteration. 40
41 Some Remarks on Degeneracy and Alternative Optima It might seem that degeneracy would be rare. After all, why should we expect that the RHS would be 0 for some variable? In reality, degeneracy is incredibly common. The simplex method is not necessarily finite unless care is taken in the pivot rule. In reality, almost no one takes care, and the simplex method is not only finite, but it is incredibly efficient. The issue of alternative optima arises frequently, and is of importance in practice. 41
42 Overview The simplex method has been a huge success in optimization. It solves linear programs efficiently We can solve problems with millions of variables It can be a starting point for problems that are not linear The simplex method requires some simple techniques to get started Transformation into standard form Big M method or variant In practice it requires lots of engineering numerical stability choosing pivot rules that are fast in practice carrying out fast linear algebra CPLEX is one of several good LP codes 42
43 Summary of today s lecture The simplex method starts with an initial bfs and (under non-degeneracy) each pivot improves the objective function. It is very effective in practice. very good rules for choosing the entering variable very good implementations in practice to speed up the linear algebra Degeneracy: 0 s in RHS can lead to no improvement in the objective function. 0 s in the reduced costs can lead to alternative optima An artificial variable with a large negative cost can be used to create the first bfs. (Or use phase 1.) 43
44 Lecture check problems Section 4.6. Problems 1 and 2 Section 4.7. Problem 4 Section 4.8. Problem 1 Section Problem 2 Section Problem 4 44
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